Question

Show that the rank of a matrix C = AB is never greater than the smaller of the rank of A and the rank of B. Can it ever be strictly less than the smaller of these two numbers?

Answer 1

Question 23819Show that the rank of a matrix $\mathrm{C}=\mathrm{AB}$ is never greater than the smaller of the rank of A and the rank of B. Can it ever be strictly less than the smaller of these two numbers?.

Solution. Since $Cx = A(Bx)$, then $\mathrm{rank}(C) \leq \mathrm{rank}(A)$, next it is known that rank of transpose matrix is equal to the rank of the original matrix, thus, $\mathrm{rank}(C') = \mathrm{rank}(C)$ and $C'x = B'(A'(x))$, thus $\mathrm{rank}(C') \leq \mathrm{rank}(B') = \mathrm{rank}(B)$, thus $\mathrm{rank}(C) \leq \mathrm{rank}(A)$, $\mathrm{rank}(C) \leq \mathrm{rank}(B)$, thus $\mathrm{rank}(C) \leq \min\{\mathrm{rank}(A), \mathrm{rank}(B)\}$.